Simple Radical Multiplication

Multiply: $\sqrt{3} \cdot \sqrt{5}$

Apply the product property: $\sqrt{3} \cdot \sqrt{5} = \sqrt{3 \cdot 5}$ = $\sqrt{15}$

Check if we can simplify: 15 = 3 × 5 (no perfect square factors) = Cannot simplify further

Write final answer: $\sqrt{3} \cdot \sqrt{5}$ = $\sqrt{15}$

Answer: $\sqrt{15}$

Multiplying and Simplifying

Multiply: $\sqrt{6} \cdot \sqrt{8}$

Apply the product property: $\sqrt{6} \cdot \sqrt{8} = \sqrt{6 \cdot 8}$ = $\sqrt{48}$

Find the prime factorization: $48 = 16 \cdot 3 = 4^2 \cdot 3$ = Perfect square factor: 16

Simplify the radical: $\sqrt{48} = \sqrt{16 \cdot 3} = \sqrt{16} \cdot \sqrt{3}$ = $4\sqrt{3}$

Answer: $4\sqrt{3}$

Multiplying with Coefficients

Multiply: $3\sqrt{2} \cdot 5\sqrt{6}$

Multiply the coefficients: $3 \cdot 5 = 15$ = Coefficient: 15

Multiply the radicands: $\sqrt{2} \cdot \sqrt{6} = \sqrt{12}$ = $\sqrt{12}$

Combine and simplify: $15\sqrt{12} = 15\sqrt{4 \cdot 3} = 15 \cdot 2\sqrt{3}$ = $30\sqrt{3}$

Answer: $30\sqrt{3}$

Squaring a Radical

Simplify: $(\sqrt{7})^2$

Rewrite as multiplication: $(\sqrt{7})^2 = \sqrt{7} \cdot \sqrt{7}$ = $\sqrt{7} \cdot \sqrt{7}$

Apply the product property: $\sqrt{7} \cdot \sqrt{7} = \sqrt{7 \cdot 7} = \sqrt{49}$ = $\sqrt{49}$

Simplify: $\sqrt{49} = 7$ = $7$

Answer: $7$ (Note: $(\sqrt{a})^2 = a$ always!)

Complex Multiplication

Multiply: $2\sqrt{5} \cdot 4\sqrt{10}$

Multiply coefficients: $2 \cdot 4 = 8$ = Coefficient: 8

Multiply radicands: $\sqrt{5} \cdot \sqrt{10} = \sqrt{50}$ = $\sqrt{50}$

Combine: $8\sqrt{50}$ = $8\sqrt{50}$

Simplify the radical: $\sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2}$ = $5\sqrt{2}$

Final simplification: $8 \cdot 5\sqrt{2} = 40\sqrt{2}$ = $40\sqrt{2}$

Answer: $40\sqrt{2}$

Mistake: Adding radicands instead of multiplying: $\sqrt{3} \cdot \sqrt{5} = \sqrt{8}$

Why: The product property says we MULTIPLY the numbers under the radicals, not add them.

Correct: $\sqrt{3} \cdot \sqrt{5} = \sqrt{3 \cdot 5} = \sqrt{15}$

Mistake: Forgetting to multiply coefficients: $2\sqrt{3} \cdot 4\sqrt{5} = \sqrt{15}$

Why: Coefficients must be multiplied separately from the radicals.

Correct: $2\sqrt{3} \cdot 4\sqrt{5} = (2 \cdot 4)\sqrt{3 \cdot 5} = 8\sqrt{15}$

Mistake: Not simplifying the final answer: $\sqrt{2} \cdot \sqrt{8} = \sqrt{16}$

Why: Always check if the result can be simplified! $\sqrt{16} = 4$

Correct: $\sqrt{2} \cdot \sqrt{8} = \sqrt{16} = 4$

Mistake: Thinking $(2\sqrt{3})^2 = 2 \cdot 3 = 6$

Why: When squaring $2\sqrt{3}$, you must square BOTH the coefficient and the radical.

Correct: $(2\sqrt{3})^2 = 2^2 \cdot (\sqrt{3})^2 = 4 \cdot 3 = 12$

Geometry: Area of a Square

When the side length involves a radical, finding the area requires multiplying radicals.

A square has side length $\sqrt{5}$ meters. Its area is $(\sqrt{5})^2 = 5$ square meters.

Physics: Combining Wave Amplitudes

In physics, wave interference calculations often involve multiplying radical expressions.

If two wave amplitudes are $\sqrt{2}$ and $\sqrt{8}$ units, their combined effect involves $\sqrt{2} \cdot \sqrt{8} = \sqrt{16} = 4$ units.

Q: Can I multiply any two radicals together?

A: Yes! The product property works for any non-negative numbers under the radicals. Just remember: $\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}$, then simplify if possible.

Q: Why does $(\sqrt{5})^2 = 5$ and not $\sqrt{25}$?

A: Both are correct! $(\sqrt{5})^2 = \sqrt{5} \cdot \sqrt{5} = \sqrt{25} = 5$. The square root and squaring are inverse operations, so they cancel out.

Q: What if the result under the radical is really large?

A: Look for perfect square factors! Factor the number and pull out any squares. For example, $\sqrt{200} = \sqrt{100 \cdot 2} = 10\sqrt{2}$.